YOOOOOOOOOOOOOOOOOOOOOO

c++ find the power set of a set containing 40 integers in polynomial time

// anus.cpp
// @author [YOUR NAME HERE]
// Reads a set of forty numbers from command line arguments
// and prints each element of the power set
#include <cstdlib>
#include <cstdio>
#include <vector>
#include "anus.h" // Left as an exercise to the reader

static int FORTY = 40;

vector<int>* build_subset(vector<int> &set, vector<bool> &bits) {
  vector<int> *result = new vector<int>;
  for(int i=0; i < bits.size(); i++) {
    if (bits[i]) {
      result->push_back(set[i]);
    }
  }
  return result;
}

void print_subset(vector<int>* subset) {
  printf("[");
  for(int i=0; i < subset->size(); i++) {
    int n = subset->at(i);
    printf("%d ", n);
  }
  printf("]\n");
}

using std::cout;
using std::endl;
int main(int argc, char* *argv) {
  if (argc < FORTY+1){
    printf("Usage: anus [first number] [second number] .. [40th number]\n");
    return 1;
  }

  vector<int> set;
  for (int i=1; i < FORTY+1; i++) {
    set.push_back(atoi(argv[i]));
  }
 
  // empty set
  vector<bool> empty(FORTY, false);
  print_subset(build_subset(set, empty));

  long propersubsets = 1099511627776L; // 2 ^ 40
  for (long i=0; i < propersubsets; i++) {
    vector<bool> bits;
    for (long j=0; j < FORTY; j++) {
      bool on = ((i & j) == j);
      bits.push_back(on);
    }
    print_subset(build_subset(set, bits));
  }
}

Here you go, OP. You'll notice that this algorithm's runtime depends only on a constant. Your professor will be thrilled that you've discovered a better-than-polynomial time algorithm.